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Math Expert V
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Question Stats: 38% (01:34) correct 63% (01:53) wrong based on 56 sessions

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For prime numbers $$x$$ and $$y$$, $$x^3*y^5=z^4$$. How many positive factors does positive integer $$z$$ have?

A. One
B. Two
C. Three
D. Four
E. Five

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5
Official Solution:

For prime numbers $$x$$ and $$y$$, $$x^3*y^5=z^4$$. How many positive factors does positive integer $$z$$ have?

A. One
B. Two
C. Three
D. Four
E. Five

Notice that $$z^4$$ is a perfect square, thus it must have odd number of factors. Since $$z^4$$ equals to $$x^3*y^5$$, then $$x^3*y^5$$ must also have odd number of factors.

Now, if $$x$$ and $$y$$ are different primes, then the number of factors of $$x^3*y^5$$ will be $$(3 + 1)(5 + 1) = 24$$, which is not odd. Therefore $$x$$ and $$y$$ MUST be the same prime.

In this case $$x^3*y^5 = x^3*x^5 = x^8 = z^4$$.

$$x^8 = z^4$$;

$$x^2 = z$$.

Since $$x$$ is prime, then the number of factors of $$z$$ is $$(2 + 1) = 3$$.

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"Notice that z4 is a perfect square, thus it must have odd number of factors. Since z4 equals to x3∗y5, then x3∗y5 must also have odd number of factors."

Question about perfect squares. Would z^6 also be a perfect square?

Also, why do x^3 and y^5 need odd factors, given z^4?
Math Expert V
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Posts: 58322

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HunterJ wrote:
"Notice that z4 is a perfect square, thus it must have odd number of factors. Since z4 equals to x3∗y5, then x3∗y5 must also have odd number of factors."

Question about perfect squares. Would z^6 also be a perfect square?

Also, why do x^3 and y^5 need odd factors, given z^4?

1. Any integer in even power is a perfect square.
2. $$z^4$$ is a perfect square. A perfect square has odd number of factors. Since $$z^4$$ equals to $$x^3*y^5$$, then $$x^3*y^5$$ (the product of x^3 and y^5, not the multiples separately) must also have odd number of factors.

Finding the Number of Factors of an Integer:

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

BACK TO YOUR QUESTION:
In case of a perfect square, say for $$6^2 = 2^2*3^2$$, the powers of primes will be always even, so when counting the number of factors we'd have $$(even + 1)(even + 1) = (2 + 1)(2 + 1) = 3*3 = odd*odd = odd$$.

Tips about the perfect square:
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.

Hope it helps.
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z cannot be a prefect square unless x=y. x and y are prime numbers, we cant get a prefect square with unequal powers of prime numbers. hence, x=y.
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I think this is a high-quality question and I agree with explanation.
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Bunuel wrote:
Official Solution:

For prime numbers $$x$$ and $$y$$, $$x^3*y^5=z^4$$. How many positive factors does positive integer $$z$$ have?

A. One
B. Two
C. Three
D. Four
E. Five

Notice that $$z^4$$ is a perfect square, thus it must have odd number of factors. Since $$z^4$$ equals to $$x^3*y^5$$, then $$x^3*y^5$$ must also have odd number of factors.

Now, if $$x$$ and $$y$$ are different primes, then the number of factors of $$x^3*y^5$$ will be $$(3 + 1)(5 + 1) = 24$$, which is not odd. Therefore $$x$$ and $$y$$ MUST be the same prime.

In this case $$x^3*y^5 = x^3*x^5 = x^8 = z^4$$.

$$x^8 = z^4$$;

$$x^2 = z$$.

Since $$x$$ is prime, then the number of factors of $$z$$ is $$(2 + 1) = 3$$.

SO Bunuel just to reiterate last line. Does that mean the factors are 1,x,z ?
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Math Expert V
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mbsingh wrote:
Bunuel wrote:
Official Solution:

For prime numbers $$x$$ and $$y$$, $$x^3*y^5=z^4$$. How many positive factors does positive integer $$z$$ have?

A. One
B. Two
C. Three
D. Four
E. Five

Notice that $$z^4$$ is a perfect square, thus it must have odd number of factors. Since $$z^4$$ equals to $$x^3*y^5$$, then $$x^3*y^5$$ must also have odd number of factors.

Now, if $$x$$ and $$y$$ are different primes, then the number of factors of $$x^3*y^5$$ will be $$(3 + 1)(5 + 1) = 24$$, which is not odd. Therefore $$x$$ and $$y$$ MUST be the same prime.

In this case $$x^3*y^5 = x^3*x^5 = x^8 = z^4$$.

$$x^8 = z^4$$;

$$x^2 = z$$.

Since $$x$$ is prime, then the number of factors of $$z$$ is $$(2 + 1) = 3$$.

SO Bunuel just to reiterate last line. Does that mean the factors are 1,x,z ?

Absolutely. Say if x = 2, then y = 4 and the factors of 4 are 1, 2, and 4.
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Bunuel wrote:
HunterJ wrote:
"Notice that z4 is a perfect square, thus it must have odd number of factors. Since z4 equals to x3∗y5, then x3∗y5 must also have odd number of factors."

Question about perfect squares. Would z^6 also be a perfect square?

Also, why do x^3 and y^5 need odd factors, given z^4?

1. Any integer in even power is a perfect square.
2. $$z^4$$ is a perfect square. A perfect square has odd number of factors. Since $$z^4$$ equals to $$x^3*y^5$$, then $$x^3*y^5$$ (the product of x^3 and y^5, not the multiples separately) must also have odd number of factors.

Finding the Number of Factors of an Integer:

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

BACK TO YOUR QUESTION:
In case of a perfect square, say for $$6^2 = 2^2*3^2$$, the powers of primes will be always even, so when counting the number of factors we'd have $$(even + 1)(even + 1) = (2 + 1)(2 + 1) = 3*3 = odd*odd = odd$$.

Tips about the perfect square:
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.

Hope it helps.

Hi Bunuel,

Considering the 4th point in the tips for perfect square (great tips btw), we can directly assume in the question that x and y are the same prime numbers since their powers in the question are odd and z is a perfect square? Thanks!
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Math Expert V
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ashikaverma13 wrote:
Bunuel wrote:
HunterJ wrote:
"Notice that z4 is a perfect square, thus it must have odd number of factors. Since z4 equals to x3∗y5, then x3∗y5 must also have odd number of factors."

Question about perfect squares. Would z^6 also be a perfect square?

Also, why do x^3 and y^5 need odd factors, given z^4?

1. Any integer in even power is a perfect square.
2. $$z^4$$ is a perfect square. A perfect square has odd number of factors. Since $$z^4$$ equals to $$x^3*y^5$$, then $$x^3*y^5$$ (the product of x^3 and y^5, not the multiples separately) must also have odd number of factors.

Finding the Number of Factors of an Integer:

First make prime factorization of an integer $$n=a^p*b^q*c^r$$, where $$a$$, $$b$$, and $$c$$ are prime factors of $$n$$ and $$p$$, $$q$$, and $$r$$ are their powers.

The number of factors of $$n$$ will be expressed by the formula $$(p+1)(q+1)(r+1)$$. NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: $$450=2^1*3^2*5^2$$

Total number of factors of 450 including 1 and 450 itself is $$(1+1)*(2+1)*(2+1)=2*3*3=18$$ factors.

BACK TO YOUR QUESTION:
In case of a perfect square, say for $$6^2 = 2^2*3^2$$, the powers of primes will be always even, so when counting the number of factors we'd have $$(even + 1)(even + 1) = (2 + 1)(2 + 1) = 3*3 = odd*odd = odd$$.

Tips about the perfect square:
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.

Hope it helps.

Hi Bunuel,

Considering the 4th point in the tips for perfect square (great tips btw), we can directly assume in the question that x and y are the same prime numbers since their powers in the question are odd and z is a perfect square? Thanks!

_______________
Yes, that's correct.
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Just curious multiplication of 2 fractions can be an integer = 2/3 x 3/2 is equal to 1. Is it not possible that x is not equal to y? we find two prime numbers such that forth root of X^3 and Y^5 yield fractions that are reciprocal to each other. Not possible???

P.S. I solved it by thinking that since the question is directly asking how many Prime factor , and because there can be only one correct ans take both X and Y same ( Try to solve with special unique cases when you can) Intern  B
Joined: 21 May 2015
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I think this is a high-quality question and I agree with explanation.
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I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. I can't think of any positive integer z which satisfies the above equation. the only way according me which satisfies the equation is when x and y are same.
can explain me with a help of an example?
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Amanrohra wrote:
For prime numbers $$x$$ and $$y$$, $$x^3*y^5=z^4$$. How many positive factors does positive integer $$z$$ have?

A. One
B. Two
C. Three
D. Four
E. Five

Notice that $$z^4$$ is a perfect square, thus it must have odd number of factors. Since $$z^4$$ equals to $$x^3*y^5$$, then $$x^3*y^5$$ must also have odd number of factors.

Now, if $$x$$ and $$y$$ are different primes, then the number of factors of $$x^3*y^5$$ will be $$(3 + 1)(5 + 1) = 24$$, which is not odd. Therefore $$x$$ and $$y$$ MUST be the same prime.

In this case $$x^3*y^5 = x^3*x^5 = x^8 = z^4$$.

$$x^8 = z^4$$;

$$x^2 = z$$.

Since $$x$$ is prime, then the number of factors of $$z$$ is $$(2 + 1) = 3$$.

I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. I can't think of any positive integer z which satisfies the above equation. the only way according me which satisfies the equation is when x and y are same.
can explain me with a help of an example?

It seems that you did not read the solution carefully.

YES, as it's clearly stated in the solution, x and y MUST be the same prime. For example consider x = y = 2. Note here that unless it is explicitly stated otherwise, different variables CAN represent the same number.
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This is a high quality question.
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Another approach would be to pick numbers, which worked for me...

$$x=3$$ and $$y=3$$, the question does not state that the prime number have to be distinct... this makes it easier to work with the same primes.

$$3^3*3^5=3^8$$ so we know that $$3^8=z^4$$, therefore $$9^4=z^4$$ --> $$z=9$$

Factors of $$z=9$$ --> 1,3,9
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I solved this Q like this:

$$x^3 * y^5 = (x*y)^3*y^2$$

Now y^2 has to be equal to x*y to make it a power of 4.

$$y^2= x*y$$ leads to x=y
Therefore same prime numbers.

So $$z= (prime^2)^2$$

Therefore number of factors = 2+1. Re: M31-22   [#permalink] 26 Feb 2019, 10:47
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